Question

Sand is pouring from a pipe at the rate of $12 cm^3/s $. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4cm?

Answer 1

Hint: In this question, get the relation between two quantities and substitute in the formula of volume. Now if y = f(x), then dy/dt measures the rate of change of y with respect to ‘t’ using this to get the required rate at which the height of the sand cone is increasing.

Complete step-by-step answer:
Let r be the radius and h be the height of the cone formed by the falling sand at any time t.
Let V be the volume of the sand-cone at time t.
${\rm{V}} = \dfrac{1}{3}{\rm{\pi }}{{\rm{r}}^2}{\rm{h}}$ …. (1)
From question statement, Height of the cone is always one-sixth of the radius
$h = \dfrac{r}{6}$
So, r = 6h
Now substitute the value of r in equation 1^st, we get
${\rm{V}} = \dfrac{1}{3}{\rm{\pi }}{\left( {6h} \right)^2}{\rm{h}}$
$ \Rightarrow {\rm{V}} = \dfrac{1}{3}{\rm{\pi }} \times 36{h^2} \times {\rm{h}}$
$ \Rightarrow {\rm{V}} = 12{\rm{\pi }} \times {{\rm{h}}^3}$
Differentiating w.r.t t on both sides we get,
$\dfrac{{{\rm{dv}}}}{{{\rm{dt}}}} = \dfrac{{{\rm{d}}\left( {12{\rm{\pi }} \times {{\rm{h}}^3}} \right){\rm{\;}}}}{{{\rm{dt}}}}$
Here 12π is constant so we can take outside from differentiating part,
$ \Rightarrow \dfrac{{{\rm{dv}}}}{{{\rm{dt}}}} = 12{\rm{\pi }} \times \dfrac{{{\rm{d}}\left( {{{\rm{h}}^3}} \right){\rm{\;}}}}{{{\rm{dt}}}}$
$ \Rightarrow \dfrac{{{\rm{dv}}}}{{{\rm{dt}}}} = 12{\rm{\pi }} \times \dfrac{{{\rm{d}}\left( {{{\rm{h}}^3}} \right){\rm{\;}}}}{{{\rm{dh}}}} \times \dfrac{{{\rm{dh}}}}{{{\rm{dt}}}}$
As we know, $\dfrac{{{\rm{d}}{{\rm{x}}^{\rm{n}}}}}{{{\rm{dx}}}} = {\rm{n}}{{\rm{x}}^{{\rm{n}} - 1}}$
$\therefore \dfrac{{{\rm{dv}}}}{{{\rm{dt}}}} = 12{\rm{\pi }} \times 3{{\rm{h}}^2} \times \dfrac{{{\rm{dh}}}}{{{\rm{dt}}}}{\rm{\;}}$
Given: Sand is pouring from a pipe at the rate of 12cm³/s
So, $\dfrac{{{\rm{dv}}}}{{{\rm{dt}}}} = 12{\rm{c}}{{\rm{m}}^3}/{\rm{sec}}$
$ \Rightarrow 12{\rm{\pi }} \times 3{{\rm{h}}^2} \times \dfrac{{{\rm{dh}}}}{{{\rm{dt}}}} = 12\;$
When h = 4,
$ \Rightarrow 12{\rm{\pi }} \times 3 \times {4^2} \times \dfrac{{{\rm{dh}}}}{{{\rm{dt}}}} = 12$
$\therefore \dfrac{{{\rm{dh}}}}{{{\rm{dt}}}} = \dfrac{1}{{48\pi }}$
Hence the height of the sand cone is increasing at the rate of $\dfrac{1}{{48{\rm{\pi }}}}$ cm/sec

Note: Whenever we face such types of questions, the first step is get the relation between the quantities given in the question. Now differentiate volume with respect to t you will get the rate at which sand is pouring. Then simplify the expression in order to get the required rate at which height of the sand cone is increasing.